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=12Y^2-11Y-5
We move all terms to the left:
-(12Y^2-11Y-5)=0
We get rid of parentheses
-12Y^2+11Y+5=0
a = -12; b = 11; c = +5;
Δ = b2-4ac
Δ = 112-4·(-12)·5
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*-12}=\frac{-30}{-24} =1+1/4 $$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*-12}=\frac{8}{-24} =-1/3 $
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